cái gì vậy bạn

? bài ở đâu

A = ( x - 3 )( x - 7 ) - ( 2x - 5 )( x - 1 )

= x² - 10x + 21 - ( 2x² - 7x + 5 )

= x² - 10x + 21 - 2x² + 7x - 5

= -x² - 3x + 16

Rồi bạn thế lần lượt x = 0 ; x = 1 ; x = -1 vô là xong (:

B = ( 3x + 5 )( 2x - 1 ) + ( x - 1 )( 3x + 2 )

= 6x² + 7x - 5 + 3x² - x - 2

= 9x² + 6x - 7 

| x | = 2 <=> x = 2 hoặc x = -2

Rồi thế vô (:

Good luck 

Đề: Tính???

2012² - 2011² + 2010² - 2009² + ... + 2² - 1²

= (2012 + 2011) (2012 - 2011) + (2010 + 2009) (2010 - 2009) + ... + (2 + 1) (2 - 1)

= 2012 + 2011 + 2010 + 2009 + ... + 2 + 1

Số số hạng là: (2012 - 1) : 1 + 1 = 2012 (số)

Tổng bằng: (1 + 2012) . 2012 : 2 = 2025078

Vậy 2012² - 2011² + 2010² - 2009² + ... + 2² - 1² = 2025078.

1 + 2xy - x² - y²

= 1 - ( x² - 2xy + y² )

= 1² - ( x - y )²

= [ 1 - ( x - y ) ][ 1 + ( x - y ) ]

= ( y - x + 1 )( x - y + 1 )

a² + b² - c² - d² - 2ab + 2cd

= ( a² - 2ab + b² ) - ( c² - 2cd + d² )

= ( a - b )² - ( c - d )²

= [ ( a - b ) - ( c - d ) ][ ( a - b ) + ( c - d ) ]

= ( a - b - c + d )( a - b + c - d )

a³b³ - 1

= ( ab )³ - 1³

= ( ab - 1 )[ ( ab )² + ab.1 + 1² ]

= ( ab - 1 )( a²b² + ab + 1 )

x²( y - z ) + y²( z - x ) + z²( x - y )

= z²( x - y ) + x²y - x²z + y²z + y²x

= z²( x - y ) + ( x²y - y²x ) - ( x²z - y²z )

= z²( x - y ) + xy( x - y ) - z( x² - y² )

= z²( x - y ) + xy( x - y ) - z( x + y )( x - y )

= ( x - y )[ z² + xy - z( x + y ) ]

= ( x - y )( z² + xy - zx - zy )

= ( x - y )[ ( z² - zx ) - ( zy - xy ) ]

= ( x - y )[ z( z - x ) - y( z - x ) ]

= ( x - y )( z - x )( z - y )

Ta có tam giác Pascal:

Bậc 0: 1

Bậc 1: 1    1

Bậc 2: 1    2    1

Bậc 3: 1    3    3    1

Bậc 4: 1    4    6    4    1

Bậc 5: 1    5   10   10   5    1

Bậc 6: 1    6   15   20  15   6    1

Bậc 7: 1    7   21   35  35   21   7    1

Bậc 8: 1    8   28   56  70   56  28   8    1

Suy ra:

      \(\left(a+b\right)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\)

      \(\left(a+b\right)^8=a^8+8a^7b+28a^6b^2+56a^5b^3+70a^4b^4+56a^3b^4+28a^2b^6+8ab^7+a^8\)

Chúc bn hok tốt ^_^

Bài 1

( x - y )² - 4

= ( x - y )² - 2²

= ( x - y - 2 )( x - y + 2 )

x² - 16( x + y )²

= x² - 4²( x + y )²

= x² - ( 4x + 4y )²

= ( x² - 4x - 4y )( x² - 4x + 4y )

8x³ + 36x²y + 54xy² + 27y³

= ( 2x )³ + 3.( 2x )².3y + 3.2x.( 3y )² + ( 3y )³

= ( 2x + 3y )³

Bài 2.

a) ( x² + 4 )( x² - 4 ) - ( x² + 1 )( x² - 1 )

= [ ( x² )² - 4² ] - [ ( x² )² - 1² ]

= x⁴ - 16 - x⁴ + 1

= -15

b) ( y - 3 )( y + 3 )( y² + 9 )( y² + 2 )( y² - 2 )

= [ ( y - 3 )( y + 3 )( y² + 9 ) ][ ( y² + 2 )( y² - 2 ) ]

= { [ ( y - 3 )( y + 3 ) ]( y² + 9 ) }[ ( y² )² - 2² ]

= [ ( y² - 9 )( y² + 9 ) ]( y⁴ - 4 )

= ( y⁴ - 81 )( y⁴ - 4 )

= y⁴( y⁴ - 4 ) - 81( y⁴ - 4 )

= y⁸ - 4y⁴ - 81y⁴ + 324

= y⁸ - 85y⁴ + 324

\(a,\left(a+b\right)\left(a+b\right)=a^2+2ab+b^2\)

\(b,\left(a+2\right)\left(a+2\right)=a^2+4a+4\)

\(c,\left(a-b\right)\left(a-b\right)=a^2-2ab+b^2\)

\(d,\left(7-2y\right)\left(7-2y\right)=49-28y+4y^2\)

( a + b )( a + b ) = a( a + b ) + b( a + b ) = a² + ab + ab + b² = a² + 2ab + b²

( a + 2 )( a + 2 ) = a( a + 2 ) + 2( a + 2 ) = a² + 2a + 2a + 4 = a² + 4a + 4

( a - b )( a - b ) = a( a - b ) - b( a - b ) = a² - ab - ab + b² = a² - 2ab + b²

( 7 - 2y )( 7 - 2y ) = 7( 7 - 2y ) - 2y( 7 - 2y ) = 49 - 14y - 14y + 4y² = 49 - 28y + 4y²

( 2 - x )( x + 2 ) = 2( x + 2 ) - x( x + 2 ) = 2x + 4 - x² - 2x = 4 - x²

\(\left(\sqrt{2x}-y\right)^2=\left(\sqrt{2x}\right)^2-2\sqrt{2x}y+y^2=2x-2y\sqrt{2x}+y^2\)

\(\left(x^2-4\right)\left(x^2+4\right)=\left(x^2\right)^2-4^2=x^4-16\)

Bài 1.

a) 2x² + 3( x - 1 )( x + 1 ) - 5x( x + 1 )

= 2x² + 3( x² - 1 ) - 5x² - 5x

= 2x² + 3x² - 3 - 5x² - 5x

= -5x - 3 

b) 4( x - 1 )( x + 5 ) - ( x - 2 )( x + 5 ) - 3( x - 1 )( x + 2 )

= 4( x² + 4x - 5 ) - ( x² + 3x - 10 ) - 3( x² + x - 2 )

= 4x² + 16x - 20 - x² - 3x + 10 - 3x² - 3x + 6

= 10x - 4

Bài 2.

a) ( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) = 0

<=> -5x² - 2x + 16 + 4( x² - x - 2 ) + 2( x² - 4 ) = 0

<=> -5x² - 2x + 16 + 4x² - 4x - 8 + 2x² - 8 = 0

<=> x² - 6x = 0

<=> x( x - 6 ) = 0

<=> x = 0 hoặc x = 6

b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 0

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 0

<=> x² + 5x + 6 - x² - 3x + 10 = 0

<=> 2x + 16 = 0

<=> 2x = -16

<=> x = -8

Bài 3.

A = ( n² + 3n - 1 )( n + 2 ) - n³ + 2

= n³ + 2n² + 3n² + 6n - n - 2 - n³ + 2

= 5n² + 5n

= 5n( n + 1 ) chia hết cho 5 ( đpcm )

B = ( 6n + 1 )( n + 5 ) - ( 3n + 5 )( 2n - 1 )

= 6n² + 30n + n + 5 - ( 6n² - 3n + 10n - 5 )

= 6n² + 31n + 5 - 6n² - 7n + 5

= 24n + 10

= 2( 12n + 5 ) chia hết cho 2 ( đpcm )

bài 1:a,\(2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)

\(=2x^2+3x^2-3-5x^2-5x\)

\(=-3-5x\)

b.\(4\left(x-1\right)\left(x+5\right)-\left(x-2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)

\(=4\left(x^2+4x-5\right)-\left(x^2+3x-10\right)-3\left(x^2+x-2\right)\)

\(=4x^2+16x-20-x^2-3x+10-3x^2-3x+6\)

\(=10x-4\)

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2+2x-2x-4\right)=0\)

\(-2x+16-5x^2+4x^2-4x-8+2x^2-8=0\)

\(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 15

= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 15

= ( x² + 5x + 4 )( x² + 5x + 6 ) - 15 (*)

Đặt t = x² + 5x + 4 

(*) trở thành

t( t + 2 ) - 15

= t² + 2t - 15

= t² - 3t + 5t - 15

= t( t - 3 ) + 5( t - 3 )

= ( t - 3 )( t + 5 )

= ( x² + 5x + 4 - 3 )( x² + 5x + 4 + 5 )

= ( x² + 5x + 1 )( x² + 5x + 9 )

b) ( x + 2 )( x + 3 )²( x + 4 ) - 12

= [ ( x + 2 )( x + 4 ) ]( x + 3 )² - 12

= ( x² + 6x + 8 )( x² + 6x + 9 ) - 12 (*)

Đặt t = x² + 6x + 8

(*) trở thành

t( t + 1 ) - 12

= t² + t - 12

= t² - 3t + 4t - 12

= t( t - 3 ) + 4( t - 3 )

= ( t - 3 )( t + 4 )

= ( x² + 6x + 8 - 3 )( x² + 6x + 8 + 4 )

= ( x² + 6x + 5 )( x² + 6x + 12 )

= ( x² + x + 5x + 5 )( x² + 6x + 12 )

= [ x( x + 1 ) + 5( x + 1 ) ]( x² + 6x + 12 )

= ( x + 1 )( x + 5 )( x² + 6x + 12 )

a, Gọi\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

                \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt\(y=x^2+5x+4\)

\(\Rightarrow A=y\left(y+2\right)-15\)

        \(=y^2+2y-15\)

        \(=\left(x-3\right)\left(x+5\right)\)

Hay\(A=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

Vậy...

b,Gọi\(B=\left(x+2\right)\left(x+3\right)^2\left(x+4\right)-12\)

           \(=\left(x^2+6x+8\right)\left(x^2+6x+9\right)-12\)

Đặt\(z=x^2+6x+8\)

\(\Rightarrow B=z\left(z+1\right)-12\)

        \(=z^2+z-12\)

        \(=\left(z-3\right)\left(z+4\right)\)

Hay\(B=\left(x^2+6x+5\right)\left(x^2+6x+12\right)\)

Vậy...

Linz